Question

A survey conducted by the American Automobile Association showed that a family of four spends an average of $315.60 per day when on vacation. Assume that $315.60 is the population mean expenditure per day for a family of four and that $85.00 is the population standard deviation. Suppose that a random sample of 40 families will be selected for further study.

1. What is the standard deviation of the sampling distribution of xbar values where xbar is the mean expenditure per day for families of four? Answer

2. What is the probability that a simple random sample of the 40 families will provide a sample mean that is within $20 of the population mean?

Answer 1

We have given:

Population mean = 315.60 , Population standard deviation = 85 and sample size n=40

a) We have asked to find the standard deviation of the sampling distribution of x bar values where x bar is the mean expenditure per day for families of four.

By Central limit theorem, we know that-

We know that If X is a random variable having mean and standard deviation , let x1, x2, x3, xn....... be the random sample taken from X then the distribution of sample mean X bar follows Normal distribution with mean and standard deviation /

Hence we get -

The standard deviation of the sampling distribution of xbar values where xbar is the mean expenditure per day for families of four is  85/ sqrt(40) = 13.4397

b) We have to find P( x bar 20)

Hence P( x bar 20 ) = P( z ( 20 - 315.60) / 13.4397)

= P( z -21.995)

= 0.0000

Hope this will help you. Thank you :)